Integration by Parts Examples: Choose u and Check Each Answer

Consider $\int x e^{2x}\,dx$. The power rule cannot integrate the whole product, and integrating $e^{2x}$ alone leaves the $x$ behind. Integration by parts gives the product a useful direction: differentiate the part that becomes simpler, integrate the part that stays manageable, and then handle the smaller integral that remains.

This lesson works through integration by parts examples, from a single product to repeated use and definite bounds. If you want the meaning of an integral before the technique, start with our integral guide. The derivative guide also helps because differentiating your answer is the best check here.

Where the formula comes from

The product rule says $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Integrate both sides and move one term across. This gives the integration by parts formula described in OpenStax Calculus, Volume 2:

$$\int u\,dv=uv-\int v\,du.$$

Here $u$ is the expression you choose to differentiate, and $dv$ is what you choose to integrate. After finding $du$ and $v$, the new integral $\int v\,du$ should be easier than the original one. If it is worse, reconsider the split. The formula is an algebraic consequence of the product rule, not a recipe that guarantees every split will help.

For an indefinite integral, remember the constant $C$ at the end. The choice of an antiderivative for $v$ can absorb a constant, but it is simpler to pick one convenient $v$ and add $C$ to the final answer.

Choose u and dv with a purpose

The useful question is: “Which factor shrinks when differentiated?” A polynomial such as $x^2$ becomes $2x$, then $2$, then zero. A logarithm becomes $1/x$, which often combines well with the rest. Meanwhile, exponentials and sine or cosine remain easy to integrate repeatedly.

The familiar LIATE mnemonic orders logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential factors as candidates for $u$. OpenStax presents it as a heuristic, not a theorem. Check the resulting $\int v\,du$ instead of trusting the first letter you see. A different split may be shorter, and some products are better handled by substitution.

For example, $\int 2x\cos(x^2)\,dx$ contains an inner function $x^2$ and its derivative $2x$. Set $t=x^2$. Then $dt=2x\,dx$ and the integral is $\sin(x^2)+C$. Using integration by parts would turn a simple composition into unnecessary work. In contrast, $x e^{2x}$ has no such direct inner-derivative match, while differentiating $x$ removes it.

Write a small two-column table before calculating: $u$, $du$ on one side, $dv$, $v$ on the other. Then ask whether multiplying $v$ and $du$ creates a simpler integral. This check prevents the common mistake of choosing the easy-to-integrate factor as $u$ merely because it looks familiar.

Example 1: a linear factor times an exponential

Find $\int x e^{2x}\,dx$. Choose $u=x$ so $du=dx$. Choose $dv=e^{2x}\,dx$, giving $v=\frac12e^{2x}$. The factor $\frac12$ matters: differentiating $e^{2x}$ produces an extra two.

$$\begin{aligned} \int x e^{2x}\,dx &=x\left(\frac12e^{2x}\right)-\int\frac12e^{2x}\,dx\\ &=\frac{x e^{2x}}2-\frac{e^{2x}}4+C\\ &=\frac{e^{2x}}4(2x-1)+C. \end{aligned}$$

Differentiate the factored answer. The product rule gives $\frac14[2e^{2x}(2x-1)+2e^{2x}]$, which simplifies to $x e^{2x}$. That restores the integrand exactly. If the check had produced $2x e^{2x}$, the likely error would be omitting the $\frac12$ when integrating $e^{2x}$.

The choice also has a visible stopping point: differentiating $x$ leaves a constant, so the remaining integral contains only the exponential. That is what “simpler” means in this example.

Example 2: integrate a logarithm

Find $\int\ln x\,dx$ for $x>0$. There is no written product, but you can supply a factor of one: $\ln x=1\cdot\ln x$. Choose $u=\ln x$ and $dv=dx$. Then $du=\frac1x\,dx$ and $v=x$.

$$\begin{aligned} \int\ln x\,dx &=x\ln x-\int x\frac1x\,dx\\ &=x\ln x-x+C. \end{aligned}$$

The check is short: $\frac{d}{dx}(x\ln x-x)=\ln x+1-1=\ln x$. The domain $x>0$ lets us write $\ln x$ and $1/x$ without crossing zero. The more general antiderivative $x\ln|x|-x+C$ works on intervals that do not include zero, but the stated problem stays on the positive side.

If you chose $u=1$ and $dv=\ln x\,dx$, you would need to know the original integral just to find $v$. That circular split explains why the logarithm belongs in $u$ here.

Example 3: repeat the method until the polynomial disappears

Find $I=\int x^2\sin x\,dx$. Differentiating $x^2$ reduces its degree, while integrating sine remains straightforward. First choose $u=x^2$ and $dv=\sin x\,dx$, so $du=2x\,dx$ and $v=-\cos x$:

$$I=-x^2\cos x+2\int x\cos x\,dx.$$

The remaining integral still contains a product. Let $J=\int x\cos x\,dx$ and apply parts again: $u=x$, $du=dx$, $dv=\cos x\,dx$, $v=\sin x$. Then

$$J=x\sin x-\int\sin x\,dx=x\sin x+\cos x.$$

Substitute that result into $I$:

$$\int x^2\sin x\,dx=-x^2\cos x+2x\sin x+2\cos x+C.$$

Now differentiate term by term. The derivative of $-x^2\cos x$ is $-2x\cos x+x^2\sin x$. The derivative of $2x\sin x$ is $2\sin x+2x\cos x$; that of $2\cos x$ is $-2\sin x$. All extra terms cancel, leaving $x^2\sin x$. This cancellation is a strong check on both the sign from integrating sine and the factor of two from differentiating $x^2$.

Repeated integration by parts is useful when each application makes a factor smaller. If the new integral resembles the original instead, it may still be solvable by collecting the original integral on one side, but do not keep applying the formula without a clear end point.

Example 4: keep the bounds in a definite integral

Evaluate $\int_0^1 x\ln(1+x)\,dx$. The factor $x$ is easy to integrate, but differentiating $\ln(1+x)$ produces $1/(1+x)$. Choose $u=\ln(1+x)$ and $dv=x\,dx$. Then $du=\frac{1}{1+x}\,dx$ and $v=x^2/2$. The definite form of integration by parts is $\int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du$.

$$\begin{aligned} \int_0^1 x\ln(1+x)\,dx &=\left[\frac{x^2}{2}\ln(1+x)\right]_0^1 -\frac12\int_0^1\frac{x^2}{1+x}\,dx\\ &=\frac12\ln2-\frac12\int_0^1\left(x-1+\frac1{x+1}\right)dx. \end{aligned}$$

Polynomial division gives $x^2/(x+1)=x-1+1/(x+1)$. Its integral from zero to one is $1/2-1+\ln2=\ln2-1/2$. Therefore

$$\int_0^1 x\ln(1+x)\,dx =\frac12\ln2-\frac12\left(\ln2-\frac12\right)=\frac14.$$

This positive result is plausible because $x\ln(1+x)$ is nonnegative on $[0,1]$. As an independent algebra check, an antiderivative is $\frac{x^2-1}{2}\ln(1+x)-\frac{x^2}{4}+\frac{x}{2}$; evaluating at one and zero also gives $1/4$. For definite integrals, carry the same endpoints through the new integral. There is no $+C$ in the final number.

Three practice problems with solutions

1. A polynomial and an exponential. Find $\int 3x e^x\,dx$.

Take $u=3x$ and $dv=e^x\,dx$. Then $du=3\,dx$ and $v=e^x$, so

$$\int 3x e^x\,dx=3x e^x-3\int e^x\,dx=3e^x(x-1)+C.$$

Differentiate $3e^x(x-1)$: it gives $3e^x(x-1)+3e^x=3xe^x$. The result matches the original integrand.

2. A logarithm with a coefficient. Find $\int x\ln x\,dx$ for $x>0$.

Choose $u=\ln x$ and $dv=x\,dx$. Then $du=dx/x$ and $v=x^2/2$:

$$\int x\ln x\,dx=\frac{x^2}{2}\ln x-\frac12\int x\,dx =\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.$$

Differentiation gives $x\ln x+x/2-x/2=x\ln x$. Keeping the two $x/2$ terms visible makes the check easier to follow.

3. A definite integral. Evaluate $\int_0^{\pi/2}x\cos x\,dx$.

Take $u=x$ and $dv=\cos x\,dx$, giving $du=dx$ and $v=\sin x$. Apply the bounds after the parts step:

$$\begin{aligned} \int_0^{\pi/2}x\cos x\,dx &=[x\sin x]_0^{\pi/2}-\int_0^{\pi/2}\sin x\,dx\\ &=\frac\pi2-[-\cos x]_0^{\pi/2}\\ &=\frac\pi2-1. \end{aligned}$$

Both $x$ and $\cos x$ are nonnegative on this interval, and $\pi/2-1$ is positive. Differentiating $x\sin x+\cos x$ gives $x\cos x$, which checks the antiderivative used to evaluate the bounds.

To practise the underlying antiderivative choices, use Math Zen's integrals topic. For a later application of repeated integration methods, see the Laplace transforms guide.